[Leetcode] 365. Water and Jug Problem 解题报告_r supply available. you need to determine whether -优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/77511178

本文探讨了使用两个容量分别为x和y升的桶精确测量z升水的问题。通过数学方法和辗转相除法求最大公约数，给出了判断能否实现精确测量的充要条件，并提供了简洁的C++代码实现。

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题目：

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

Fill any of the jugs completely with water.
Empty any of the jugs.
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example 1: (From the famous "Die Hard" example)

Input: x = 3, y = 5, z = 4
Output: True

Example 2:

Input: x = 2, y = 6, z = 5
Output: False

思路：

这是一道数学题目（不过顺便也考查了一下求最大公约数的辗转相除法^_^），具体解释在youtube上面可以看到： "Die Hard" example，我自己也没明白具体原理。不过总体来讲要用x升和y升大小的两个桶精确量出z升水，其充要条件是：1）z必须小于等于（x + y）（否则装不下）；2）z必须可以除尽x和y的最大公约数。其中条件1）显而易见；但是条件2）是为什么呢？求解释。

代码：

class Solution {
public:
    bool canMeasureWater(int x, int y, int z) {
        if(z > x + y) {
            return false;
        }
        return z == 0 ? true : z % gcd(x, y) == 0;
    }
private:
    int gcd(int x, int y) {  
        return y == 0 ? x : gcd(y, x % y);
    }
};